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antoniya [11.8K]
2 years ago
13

Give an example of intense aerobics activity. Prompt must be accurate. ​

Physics
1 answer:
Lubov Fominskaja [6]2 years ago
7 0

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

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The windowpanes are___________ a. opaque b. transparent c. absorbent .
kogti [31]

Answer:

The windowpanes are- transparent.

The color of the panes are due to the wavelengths of light that the glass- allows to pass through

Explanation:

Just answered the question.

4 0
3 years ago
Read 2 more answers
Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal
elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

  • W = F║d
  • where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

  • W = (87 N)(4.5 m)
  • W = 391.5 J  

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0
2 years ago
Read 2 more answers
Answer the following question​
Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
2 years ago
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t
Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

                              = \dfrac{2\pi }{1.5}  

                              = 4.189 rad/s

Angular acceleration α = \dfrac{\omega}{t}

                                      = \dfrac{4.189}{40}

                                      = 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

       v = 2πr/T

          = \dfrac{2\pi \times 6.2}{1.5}

          = 25.97 m/s

So centripetal acceleration.

        a = \dfrac{v^2}{r}

          = \dfrac{25.97^2}{6.2}

          =  108.781 m/s^2

          = 11.1 g    

in combination with the gravitation acceleration.

a_{total} = \sqrt{(11.1g)^2+g^2}

a_{total}= 11.145 g

6 0
3 years ago
A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis
CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car v=10 m/s

mass of clump m=0.5 kg

Radius of car tire r=0.2 m

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

F_c=\frac{mv^2}{r}

where m=mass\ of\ element

v=speed

r=distance\ from\ center

F_c=\frac{0.5\times 10^2}{0.2}

F_c=250\ N (inward)

           

3 0
3 years ago
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